Shikamaru's Maths Challange

*This challenge was put up on 14 June 2004 and solved on 21 June 2004 with 6 answers, 3 of which are correct, and 3 wrong ones.

Challenge #03 - Count the Ninjas -solved-

On your way to the training ground, you see a group of genins fighting. It's chaotic, you can't tell who's fighting who and you can't count how many ninjas there are. Your fists still hurt from beating up dishonest shopkeepers so you can't break up the fight either.

Prove that at least 2 ninjas are fighting the same number of ninjas.

*Answer this Maths question correctly and enter our Hall of Fame! Correct answers includes the correct reasoning. The answer will be revealed when someone finally gets the answer correct. You can try more than once.

(21/06) Joshua Crammer : If there are x ninjas, than each person, fighting a different number of people, will follow a number in this pattern: 1,2,3…x-2,x-1,x. However, since the x th person would have to fight x people, that person would have to fight himself. That of course is impossible, and so that person must pick another number in that series. That number has already been picked by someone else, and thus we come to the conclusion that two people are fighting the same number of people.

Shika : You're right ! Maybe I should make this more challenging.

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(22/06) Cecil Artavion : The idea is this:

Let's say you have 10 boxes and 11 oranges. If you place oranges in boxes, at least 1 box must have 2 oranges.

If you have more objects than boxes, one of the boxes contains at least 2 objects. It seems obvious (and I guess it is), but the consequences are astounding.

For example, in this problem: The boxes are the number of kids a particular kid is fighting. The objects are the kids themselves. The boxes are labeled 1 to N-1 (where N is the total number of kids). If kid A is fighting 3 kids, he gets placed in box 3. If kid B is fighting 5 kids, he is placed in box 5.

There are only N-1 boxes because there are N kids. It's possible for one of them to fight all the others, which is N-1 people the kid is fighting. But that's the maximum number of people the kid could be fighting.

The idea is that there are more kids than boxes, so two kids must be placed in the same box which means they are fighting the same number of kids (even if they aren't fighting the same kids).

Shika : Congratulations! You're right again!

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(24/06) Warbringer : Ok, sorry for making you read all that crap, it wasn't very brilliant. Guess I was bored... :)

Here goes :

Let's suppose that there aren't at least 2 ninjas that are fighting the same number of ninjas, which means that none is fighting the same number.

There are N ninjas. Each of them can fight against a maximum of N-1 ninjas and a minumum of 1. That makes N-1 choices for N ninjas, which will obviously lead to a duplicate.